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The colour of a complex compound is due to unpaired electrons. As per crystal field theory, $\ce{K4[Fe(CN)6]}$ has no unpaired electrons so it has to be colourless. But then why is it coloured?

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Jan
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JM97
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    Color is due to electron transitions. Whether or not the electrons were paired before the transition is less important. – Ivan Neretin May 01 '16 at 06:48

1 Answers1

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There is no direct connection between having unpaired electrons and having a colour. In fact, many colourful compounds have no unpaired electrons and yet display a bright colour: potassium permanganate, potassium dichromate, azobenzene, sudan red, phenolphthalein, iodine and many, many more. Arguably, there are more coloured compounds known in the singlet state (which means no unpaired spins in the vast majority of cases) than in the dublet, triplet or any other state.

So what causes the colour? It is the possibility of electron transitions from a lower to a higher orbital. If that transition is equivalent to a photon with a visible wavelength, we observe a colour. Examples:

  • Most phenyl groups absorb light at c. $\pu{250 nm}$ — to short a wavelength to be visible; they are ‘ultravioletly coloured’.

  • The electronic transitions of most transition metal complexes (except $\mathrm{d^0}$, $\mathrm{d^5}$ and $\mathrm{d^{10}}$ complexes) are well within the visible range of $400$ to $\pu{700 nm}$.

  • In rare cases, extensive electronic systems may even allow the absorption maximum to be shifted outside of the visible range into the infrared range. Chlorophyll comes rather close to this, absorbing red light (the longest visible wavelengths).

For potassium hexacyanidoferrate(II), the colour we observe is yellow so the absorption must be the complementary blue/violet, indicating a relatively large energy difference.

Jan
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    Some technicalities: "no unpaired electrons" is not synonymous with "singlet state" (usually it is, but not always). For example: the $^1\Sigma_\mathrm{g}^+$ excited state of dioxygen is a singlet state but has two unpaired electrons. Nor is "one or more unpaired electrons" synonymous with the triplet state. Free sodium atoms have one unpaired electron, but the ground state is a doublet ($^2S_{1/2}$) state. – orthocresol May 01 '16 at 13:09
  • @orthocresol Thanks, I was aware that unpaired $\ne$ triplet, but the organic chemist in me pays more attention to triplet than any other unpaired state, so I just … forgot that ;) Is writing unpaired spins technically correct? – Jan May 01 '16 at 13:18
  • Hmm, I guess so. – orthocresol May 01 '16 at 13:22
  • Just wondering: Laporte and spin selection rules should have prevented the transitions, right? – William R. Ebenezer Mar 23 '20 at 11:56
  • @WilliamR.Ebenezer The Laporte rule indeed should prevent a $\mathrm{t_{2g}\rightarrow e_g^}$ transition, that is correct. However, that transition is not spin-forbidden as all $\mathrm{e_g^}$ orbitals are empty. Which transition is actually responsible for the colour is a good question because right now I am at a loss. – Jan Mar 25 '20 at 02:11
  • Actually, thinking again and looking at the salt, the colour of hexacyanidoferrate(II) is not intense so the Laporte-forbidden $\mathrm{d \rightarrow d}$ transition seems fine. Hexacyanidoferrate(III) should have another transition (possible LMCT) that explains its much more intense red colour. – Jan Mar 25 '20 at 02:21
  • @Jan, I don't understand why the spin selection rule does not apply. From what I have been taught, I understand that change in spin multiplicity is forbidden, which seems to be the case here. And regarding Laporte, yes, I agree that the color is pale and that does make sense. But again, I feel (intuitively) that LMCT might not be very significant, as I have usually seen it happening only at high metal center oxidation states. Thanks for your reply, BTW. – William R. Ebenezer Mar 26 '20 at 05:44
  • @WilliamR.Ebenezer Look at the rightmost diagram in the question and tell me which of the electrons in the $\mathrm{t_{2g}}$ orbitals are prevented from being excited to $\mathrm{e_g^}$ by the $\mathrm{e_g^}$ orbitals being completely occupied with an electron of the same spin. (I write $\mathrm{e_g^*}$ because they happen to be antibonding orbitals according to MO theory but describing them as simply $\mathrm{e_g}$ as in OP’s diagram is fine and leads to the same result.) – Jan Mar 28 '20 at 10:53
  • Of course, hexacyanidoferrate(II) looks unlikely to allow any LMCT excitations; the $\mathrm{t_{2g}}$ orbitals are fully populated so any LMCT must go directly to $\mathrm{e_g^}$ which is a very high-energy difference. However, if you remove one of the $\mathrm{t_{2g}}$ electrons (oxidation to hexacyanidoferrate(III)), this opens up space for an excitation from a lower electron level into* $\mathrm{t_{2g}}$. Thus, hexacyanidoferrate(III) looks like an LMCT transition is much more possible. – Jan Mar 28 '20 at 10:56